Answer: minimum speed of launch must be 7.45m/s
Explanation:
Given the following:
Height or distance (s) = 2.83m
The final velocity(Vf) at maximum height = 0
Upward motion, acceleration due to gravity(g) us negative = -9.8m/s^2
From the 3rd equation of motion:
V^2 = u^2 - 2gs
Where V = final velocity
u = initial velocity
Therefore, u = Vi
u = √Vf^2 - 2gs
u = √0^2 - 2(-9.8)(2.83)
u = √0 + 55.468
u = √55.468
u = 7.4476 m/s
u = 7.45m/s
Explanation:
Formula for calculating the area of a rectangle A = Length *width
For statement A;
Given area of a rectangle with measured length = 2.536 mm and width = 1.4 mm.
Area of the rectangle = 2.536mm * 1.4mm
Area of the rectangle = 3.5504mm²
The rule of significant figures states that we should always convert the answer to the least number of significant figure amount the given value in question. Since 1.4mm has 2 significant figure, hence we will convert our answer to 2 significant figure.
Area of the rectangle = 3.6mm² (to 2sf)
For statement B;
Given area of a rectangle with measured length = 2.536 mm and width = 1.41 mm.
Area of the rectangle = 2.536mm * 1.41mm
Area of the rectangle = 3.57576mm²
Similarly, Since 1.41mm has 3 significant figure compare to 2.536 that has 4sf, hence we will convert our answer to 3 significant figure.
Area of the rectangle = 3.58mm² (to 3sf)
Based on the conversion, it can be seen that 3.6mm² is greater than 3.58mm², hence the area of rectangle in statement A is greater than the area of the rectangle in statement B.
Answer:
The depth of the well, s = 54.66 m
Given:
time, t = 3.5 s
speed of sound in air, v = 343 m/s
Solution:
By using second equation of motion for the distance traveled by the stone when dropped into a well:

Since, the stone is dropped, its initial velocity, u = 0 m/s
and acceleration is due to gravity only, the above eqn can be written as:

(1)
Now, when the sound inside the well travels back, the distance covered,s is given by:

(2)
Now, total time taken by the sound to travel:
t = t' + t''
t'' = 3.5 - t' (3)
Using eqn (2) and (3):
s = 343(3.5 - t') (4)
from eqn (1) and (4):
Solving the above quadratic eqn:
t' = 3.34 s
Now, substituting t' = 3.34 s in eqn (2)
s = 54.66 m
mass gram, time sec, temp kelvin, vol liter, dens grams/cm3
Answer:
umm section 2 and 4
Explanation:
because at section 2 it starts and at section 4 it moves again and stops at 3.